Problem 7

Exercise 4.5 (Two Contour Integrals)   Using complex analysis, evaluate the integrals

$$I_1 = \int_0^\infty \frac{1-\cos x}{x^2}dx, \quad I_2 = \int_0^{2\pi}\frac{1}{2+\cos\theta}d\theta.$$ (4.15)

Proof. The first integral can be evaluated as in Fall 2019 Exercise 8 by taking

$$\int_0^\infty \frac{1-\cos x}{x^2} = \frac{1}{2}\int_{-\infty}^\infty \frac{1-\cos x}{x^2} = \Re\int_{-\infty}^\infty \frac{1 - e^{iz}}{z^2}.$$ (4.16)

The residue of the integrand is found by a series about $z=0$:

$$\frac{1 - e^{iz}}{z^2} = \frac{1 - (1 + iz - z^2/2 + \cdots)}{z^2} = -\frac{i}{z} + \cdots$$ (4.17)

so that $\operatorname{Res}_{z=0}=-i$. For an appropriate contour, take a semicircular arc with a dimple at the origin. The contour integral can be split into a few integrals, most importantly the line segments and dimple

$$\int_{\text{dimple}} + \int_\epsilon^R \frac{1 - e^{iz}}{z^2} + \int_{-R}^{-\epsilon}\frac{1 - e^{iz}}{z^2}=2\pi i(-i) = 2\pi$$ (4.18)

Since the dimple `winds around' the origin one-half times,

$$\int_{\text{dimple}} = 2\pi i/2(-i)=\pi,$$ (4.19)

so that

$$\int_\epsilon^R \frac{1 - e^{iz}}{z^2} + \int_{-R}^{-\epsilon}\frac{1 - e^{iz}}{z^2}=2\pi i(-i) = \pi.$$ (4.20)

As $\epsilon\to 0$ and $R\to\infty$, we see

$$\int_{-\infty}^\infty \frac{1 - e^{iz}}{z^2} = 2\pi$$ (4.21)

The second integral is handled similarly to Winter 2021 Problem 7. Let $z=e^{i\theta}$ for $\theta\in[0,2\pi]$ so that

$$-2i\int \frac{1}{z^2+4z+1} = -2i \int \frac{1/z}{z + 4 + 1/z}$$ (4.22)

$$= -2i\int_0^{2\pi} \frac{e^{-i\theta}}{e^{i\theta}+e^{-i\theta}+4} ie^{i\theta}d\theta$$ (4.23)

$$= 2\int_0^{2\pi}\frac{1}{2\cos\theta + 4}d\theta$$ (4.24)

$$= \int_0^{2\pi}\frac{1}{\cos\theta + 2}d\theta=$$ (4.25)

The first integral can be evaluated with the residue theorem. Identify the poles by solving

$$z^2 + 4z + 1 = 0 \iff z = \frac{-4 \pm \sqrt{16 - 4}}{2} = -2 \pm \sqrt{3}.$$ (4.26)

Only the pole at $z=-2+\sqrt{3}$ lies inside the contour of integration, so the residue here is the only one we need to compute, as follows:

$$\operatorname{Res} = \lim_{z\to-2+\sqrt{3}} \frac{z - (-2+\sqrt{3... ...} = \lim \frac{1}{2z + 4} = \frac{1}{2(-2+\sqrt{3}) + 4} = \frac{1}{2\sqrt{3}}.$$ (4.27)

Therefore,

$$\int_0^{2\pi} \frac{1}{2+\cos\theta}d\theta = -2i\int = -2i(2\pi i/2\sqrt{3})=\frac{2\pi}{\sqrt{3}}.$$ (4.28)

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